Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, r\neq 0$. $\dfrac{{(k^{4})^{-5}}}{{(k^{-4}r)^{2}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{4}}$ to the exponent ${-5}$ . Now ${4 \times -5 = -20}$ , so ${(k^{4})^{-5} = k^{-20}}$ In the denominator, we can use the distributive property of exponents. ${(k^{-4}r)^{2} = (k^{-4})^{2}(r)^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{4})^{-5}}}{{(k^{-4}r)^{2}}} = \dfrac{{k^{-20}}}{{k^{-8}r^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-20}}}{{k^{-8}r^{2}}} = \dfrac{{k^{-20}}}{{k^{-8}}} \cdot \dfrac{{1}}{{r^{2}}} = k^{{-20} - {(-8)}} \cdot r^{- {2}} = k^{-12}r^{-2}$.